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Manager
Joined: 29 May 2007
Posts: 96

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12 Aug 2007, 12:58
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction.
Of the 5 cars, 3 are red, 1 is blue, and 1 is yellow. If the cars are identical except for color, how
many different display arrangements of the 5 cars are possible?

A. 20

B. 25

C. 40

D. 60

E.125

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Director
Joined: 01 May 2007
Posts: 794

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12 Aug 2007, 13:48
I'm thinking 20. My thought process was

red, red, red, blue, yellow.

so...

5! /(3!)(1!)(1!) = 20.

I might be way off though.

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Manager
Joined: 07 Feb 2007
Posts: 54

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12 Aug 2007, 17:48
jimmyjamesdonkey wrote:
I'm thinking 20. My thought process was

red, red, red, blue, yellow.

so...

5! /(3!)(1!)(1!) = 20.

I might be way off though.

Agree with jimmy's explanation and answer. What's the OA?

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Director
Joined: 30 Jun 2007
Posts: 781

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12 Aug 2007, 19:28
Let A,B,C,D and E are filve parking slots:
A B C D E
R R R Y B

Anagram: RRRYB

A B C D E
R B Y R R

Anagram: RBYRR

Number of combinations = 5! / 3! 1! 1!
= 5 * 4
= 20

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Intern
Joined: 07 Aug 2007
Posts: 7

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14 Aug 2007, 05:06
5!/3!*1!*1!.

What's the OA?

Good question btw.

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